Question

Find an equation for a circle satisfying the given conditions. (a) Center (-1, 4), passes through (3, 7) (b) The points (7, 13) and (-3, -11) are at the ends of a diameter.

Answer 1

Answer:

Step-by-step explanation:

In order to find the equations we need the circle's general equation:

$(x-h)^{2}+(y-k)^{2}=r^2$ where:

(h,k) is the center and 'r' is the radius.

A. Because the center is (-1,4) then h=-1 and k=4.

Now we can find the radius as:

$distance=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

$distance=\sqrt{(3-(-1))^{2}+(7-4)^{2}}$

$distance=5$ so we have r=5

Then the equation is $(x+1)^{2}+(y-4)^{2}=25$

B. Because we have two points defining a diameter we can find the radius as follows:

$diameter=\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

$diameter=\sqrt{(7-(-3))^{2}+(13-(-11))^{2}}$

$diameter=26$

$radius=26/2=13$

Now let's find the center of the circle as follows:

$C=(\frac{x1+x2}{2} , \frac{y1+y2}{2})$

$C=(\frac{7-3}{2} , \frac{13-11}{2})$

$C=(2,1)$

Then the equation is $(x-2)^{2}+(y-1)^{2}=169$